Problem: Simplify the following expression: $z = \dfrac{-10y^2 - 130y - 420}{y + 6} $
Explanation: First factor the polynomial in the numerator. We notice that all the terms in the numerator have a common factor of $-10$ , so we can rewrite the expression: $ z =\dfrac{-10(y^2 + 13y + 42)}{y + 6} $ Then we factor the remaining polynomial: $y^2 + {13}y + {42} $ ${6} + {7} = {13}$ ${6} \times {7} = {42}$ $ (y + {6}) (y + {7}) $ This gives us a factored expression: $\dfrac{-10(y + {6}) (y + {7})}{y + 6}$ We can divide the numerator and denominator by $(y - 6)$ on condition that $y \neq -6$ Therefore $z = -10(y + 7); y \neq -6$